Question: $-2st - 7t - 6u + 3 = -t - 3u + 1$ Solve for $s$.
Combine constant terms on the right. $-2st - 7t - 6u + {3} = -t - 3u + {1}$ $-2st - 7t - 6u = -t - 3u - {2}$ Combine $u$ terms on the right. $-2st - 7t - {6u} = -t - {3u} - 2$ $-2st - 7t = -t + {3u} - 2$ Combine $t$ terms on the right. $-2st - {7t} = -{t} + 3u - 2$ $-2st = {6t} + 3u - 2$ Isolate $s$ $-{2}s{t} = 6t + 3u - 2$ $s = \dfrac{ 6t + 3u - 2 }{ -{2t} }$ Swap the signs so the denominator isn't negative. $s = \dfrac{ -{6}t - {3}u + {2} }{ {2t} }$